Question

the half - life of a certain tranquilizer in the bloodstream is 32 hours. how long will it take for the drug to decay to 95% of the original dosage? use the exponential decay model, ( a = a_0e^{kt} ), to solve. hours (round the final answer to one decimal place as needed. round all intermediate values to four decimal places as needed.)

Explanation:

Step1: Find the value of \(k\)

When \(t = 32\), \(A=\frac{A_{0}}{2}\). Substitute into \(A = A_{0}e^{kt}\):
\(\frac{A_{0}}{2}=A_{0}e^{32k}\)
Divide both sides by \(A_{0}\): \(\frac{1}{2}=e^{32k}\)
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{32k})\)
Since \(\ln(e^{x}) = x\), we have \(-\ln(2)=32k\)
So \(k=\frac{-\ln(2)}{32}\approx\frac{- 0.6931}{32}\approx - 0.0217\)

Step2: Find the value of \(t\) when \(A = 0.95A_{0}\)

Substitute \(A = 0.95A_{0}\) and \(k=\frac{-\ln(2)}{32}\) into \(A = A_{0}e^{kt}\):
\(0.95A_{0}=A_{0}e^{\frac{-\ln(2)}{32}t}\)
Divide both sides by \(A_{0}\): \(0.95=e^{\frac{-\ln(2)}{32}t}\)
Take the natural logarithm of both sides: \(\ln(0.95)=\ln(e^{\frac{-\ln(2)}{32}t})\)
Since \(\ln(e^{x}) = x\), we have \(\ln(0.95)=\frac{-\ln(2)}{32}t\)
Solve for \(t\): \(t=\frac{32\ln(0.95)}{-\ln(2)}\)
\(\ln(0.95)\approx - 0.0513\), \(\ln(2)\approx0.6931\)
\(t=\frac{32\times(- 0.0513)}{- 0.6931}=\frac{-1.6416}{- 0.6931}\approx2.4\)

Answer:

\(2.4\)